我有一个有趣且麻烦的任务要解决。 我需要创建一个包含歌曲和其他子播放列表的播放列表(某种列表)...每个播放列表都有一个播放模式(随机,顺序等)。可以创建这样的播放列表吗? 我曾想过要破解子播放列表,然后将多余的歌曲添加到主播放列表,或者为添加到主播放列表的每首歌曲创建一个子播放列表(我不太喜欢这个想法) 它以某种方式解决了问题,但是有必要保持每个播放列表的播放模式...
例如:
主播放列表(音序模式)具有: (1)歌曲1-1 / (2)具有随机播放模式的子播放列表(歌曲2-1,歌曲2-2,歌曲2-3)/ (3)song1-2
理想的结果: (1)歌曲1-1 / (2)song2-3(开始随机子播放列表)/ (3)歌曲2-1 / (4)歌曲2-2 / (5)song1-2 /
我应该如何处理?
答案 0 :(得分:3)
由于我怀疑这是某种家庭作业,因此我仅向您提供部分实现,因此您了解如何进行操作。
创建一个抽象类PlaylistElement
,该抽象类以后可以是Song
或另一个Playlist
。
abstract class PlaylistElement {
public abstract List<Song> printSongs();
}
实施扩展Playlist
的类PlaylistElement
。
class Playlist extends PlaylistElement {
private List<PlaylistElement> elements;
private PlaybackMode playbackMode;
@Override
public List<Song> printSongs() {
if(this.playbackMode == PlaybackMode.RANDOM) {
List<Song> songs = new ArrayList<>();
List<PlaylistElement> shuffleElements = new ArrayList<>();
//Add all PlaylistElements from elements into shuffleElements
//and shuffle the shuffleElements collection
//insert your songs into the songs collection here by sequentially
//going through your
//PlaylistElements and inserting the result of their printSongs()
//implementation (e.g. in a for-loop)
return songs;
}
else if(this.playbackMode == PlaybackMode.SEQUENTIAL) {
//you can do this on your own
}
return null;
}
}
实施扩展Song
的类PlaylistElement
。
class Song extends PlaylistElement {
private String title;
private String artist;
.
.
.
@Override
public List<Song> printSongs() {
//return a List with only this Song instance inside
return Arrays.asList(new Song[] { this });
}
}
为您的播放列表播放模式创建一个枚举。
enum PlaybackMode {
SEQUENTIAL, RANDOM;
}
希望这给您一个总体思路!为了简洁起见,省略了Getters / Setters和其他重要部分。
答案 1 :(得分:1)
尽管已经有了一些答案,但我答应提供一个示例实现。首先,我们有一个公共接口Playable
,这是要为复合设计模式实现的类。
public interface Playable {
String getSongName();
}
接下来,Song
类代表一首歌曲。
public class Song implements Playable {
private String name;
public Song(String name) {
this.name = name;
}
@Override
public String getSongName() {
return name;
}
}
为Playlist
类做准备,一个枚举代表差异播放模式。
public enum PlayingMode {
SEQUENCE, RANDOM
}
现在,最后是播放列表类。
public class Playlist implements Playable {
private String name;
private List<Playable> playables = new ArrayList<>();
private PlayingMode mode;
private Playable currentItem;
private List<Playable> next = new ArrayList<>();
public Playlist(String name, PlayingMode mode) {
this.name = name;
this.mode = mode;
}
@Override
public String getSongName() {
if (playables.isEmpty()) {
return null;
}
if (currentItem == null) {
// initialize the playing songs
next.addAll(playables);
if (mode == PlayingMode.RANDOM) {
Collections.shuffle(next);
}
currentItem = next.get(0);
} else {
// if we have a playlist, play its songs first
if (currentItem instanceof Playlist) {
String candidate = currentItem.getSongName();
if (candidate != null) {
return candidate;
}
}
int index = next.indexOf(currentItem);
index++;
if (index < next.size()) {
currentItem = next.get(index);
} else {
currentItem = null;
}
}
return currentItem != null ? currentItem.getSongName() : null;
}
private void addToNext(Playable playable) {
if (currentItem == null) {
return;
}
// if the playlist is playing, add it to those list as well
if (mode == PlayingMode.SEQUENCE) {
next.add(playable);
} else if (mode == PlayingMode.RANDOM) {
int currentIndex = next.indexOf(currentItem);
int random = ThreadLocalRandom.current().nextInt(currentIndex, next.size());
next.add(random, playable);
}
}
public void addPlayable(Playable playable) {
Objects.requireNonNull(playable);
playables.add(playable);
addToNext(playable);
}
}
一些例子:
public static void main(String[] args) {
Song song1 = new Song("Song 1");
Song song2 = new Song("Song 2");
Playlist subPlaylist1 = new Playlist("Playlist 1", PlayingMode.RANDOM);
subPlaylist1.addPlayable(new Song("Song A"));
subPlaylist1.addPlayable(new Song("Song B"));
subPlaylist1.addPlayable(new Song("Song C"));
Song song3 = new Song("Song 3");
Playlist main = new Playlist("Main", PlayingMode.SEQUENCE);
main.addPlayable(song1);
main.addPlayable(song2);
main.addPlayable(subPlaylist1);
main.addPlayable(song3);
String songName = main.getSongName();
while (songName != null) {
System.out.println("Current song is: " + songName);
songName = main.getSongName();
}
}
可以提供输出:
Current song is: Song 1
Current song is: Song 2
Current song is: Song B
Current song is: Song A
Current song is: Song C
Current song is: Song 3
您还可以在播放时添加歌曲:
while (songName != null) {
System.out.println("Current song is: " + songName);
songName = main.getSongName();
// add songs while playing
if ("Song A".equals(songName)) {
subPlaylist1.addPlayable(new Song("Song D"));
subPlaylist1.addPlayable(new Song("Song E"));
subPlaylist1.addPlayable(new Song("Song F"));
}
}
这可能导致:
Current song is: Song 1
Current song is: Song 2
Current song is: Song B
Current song is: Song A
Current song is: Song E
Current song is: Song D
Current song is: Song F
Current song is: Song C
Current song is: Song 3
一些最后的笔记:
getIndex
方法的运行时间确实是最差的 O(n),如果播放列表中有很多歌曲,则可能会出现问题。像Collection
或Set
之类的更快的Map
可以提供更好的性能,但是实现起来会更加复杂。答案 2 :(得分:0)
方法1: 创建一个名为playlist和playlist的类,该类可以保存songIds的列表,该列表可以保存来自不同播放列表或歌曲ID的一组歌曲。
class PlayList{
List<PlayListItem> playlistItems;
}
class PlayListItem{
List<String> songIds;
}
如果您想识别通过特定子播放列表添加的歌曲集,这将对您有所帮助。但是,与方法2相比,此方法使迭代几乎没有困难
方法2: 此处,在播放列表项中避免使用该列表,因此显示播放列表时的迭代很简单。但是,要确定通过特定SubPlaylist添加的songId列表,必须进行计算。
class PlayList{
List<PlayListItem> playlistItems;
}
class PlayListItem{
String songId;
String referencePlayListId;
}